a basketball is thrown vertically up with an initial velocity

t = 6 if basketball achieves underneath

A) At what time t will the basketball strike the base?

it

t = 6t=0when the basketball is thrown

t – 6 = 0

pointsthanks the support

=>

distanceof the basketball from floor = 128feet.so, 96t – 16t2 = 128

t – 2 = 0

4 or

afterthrowing.Answer:- after

=>

it

Along basketball through floor after t sec exists bythe

– 4t- 2t + 8 = 0

theground? Step by step on which you performed this recieve life-saver

=> 16t(t – 6) =0henceeither, 16t

A)6

B)2

– 6t+ 8 = 0

Step by step on which you performed this recieve life-saver

A) the distance of basketball through groundafter t sec is

=>

6 moments of throwing the basketball

s(t) = 96t – 16t2at thetime of hitting underneath the distance of basketball

6 moments of throwing the basketball

=0 or

s(t) = 96t – 16t2at thetime of hitting underneath the distance of basketball

=0 or

2 A) At what time t will the basketball strike the base? B) for just what time t will be the basketball more than 128 base above

of96 base per second. The space s (in base) of basketball from

=> 16t2 – 96 = 0

t

=0 or

=> 16t(t – 6) =0henceeither, 16t

=>t2

=0 or

t

theground after t moments is s(t)= 96t – 16t2

afterthrowing.Answer:- after

=> (t – 4)(t – 2)

fromground = 0 feetso, 96t – 16t2 = 0

8) = 0

equation,

afterthrowing.Answer:- after

will attain underneath.

=>16(t2 – 6t+